Question

A four digit number (numbered from $0000$ to $9999$) is said to be lucky if the sum of first two digits is equal to the sum of its last two digits. If a four digit number is picked up at random, then the probability that it is lucky is

• A. $>0.065$
• B. $<0.068$
• C. $0.066$
• D. $0.067$

Answer 1

Answer: (A), (B), (D)

$>0.065$
$0.067$
$<0.068$

The total number of ways of choosing the ticket is 1000.
Let the four digits number on the ticked be $x_1{x}_2{x}_3{x}_4$.
Not that $0\le x_1+x_2\le 18$ and $=\le x_3+x_4\le 18$.
Also, the number of non-negative integral solutions of $x+y=m$ (with $0\le x,y\le 9)$ is m+1 if $0\le m\le 9$ and is $19-m$ if $10\le m\le 18$.
Thus, the number of favourable ways
$=1\times 1+2\times 2+...+10\times 10+9\times 9+8\times 8+...+1\times 1$
$=2\left\{\frac{9\times 10\times 9}{6}\right\}+100=670$
$\therefore$ Probability of require event $=\frac{670}{1000}=0.067$.
Note that $0.067>0.065$ and $0.067<\mathrm{0.068.}$